Section 1.6

Homogeneous Systems, Subspaces, and Bases

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Homogeneous Systems

A homogeneous system of m equations with n unknowns x i ( i = 1.. n ) is a system of

the form

+ ........ + = 0

+ ........ + = 0

+ ........ + = 0

.............

+ ........ + = 0

Is a homogeneous system consistent or inconsistent?

The homogeneous system of linear equations is always consistent.

The solution set of a homogeneous system is never empty. In fact, , .. always satisfy such systems. This solution is called the zero or trivial solution .

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Example 1: Consider the system:

> eq1:=x[1]+x[2]+2*x[3]=0: eq2:=2*x[1]+x[2]-3*x[3]=0: print(eq1); print(eq2);

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What is the solution set of this system?

> solveqns({eq1,eq2},{x[1],x[2],x[3]});

`Select one of the following modes`

`     1. Demonstration mode`

`     2. No-Step mode`

2;

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Example 1 represents a system with infinitely many solutions and with one free variable. ******************************************************************

Example 2 Consider the system:

> eq1:=x[1]+2*x[2]-5*x[3]+6*x[4]=0: print(eq1);

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What is the solution set of this system?

> solveqns(eq1,{x[1],x[2],x[3],x[4]});

`Select one of the following modes`

`     1. Demonstration mode`

`     2. No-Step mode`

2;

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This system has infinitely many solutions with three free variable.

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Example 3: Consider the system:

> eq1:=x[1]-2*x[2]-x[3]=0: eq2:=2*x[1]-x[2]+x[3]=0: eq3:=x[1]+3*x[2]-5*x[3]=0:

> print(eq1); print(eq2); print(eq3);

What is the solution set of this system?

> solveqns({eq1,eq2,eq3},{x[1],x[2],x[3]});

`Select one of the following modes`

`     1. Demonstration mode`

`     2. No-Step mode`

2;

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This system has the trivial solution as its unique solution.

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Examples illustrate several scenarios for homogeneous systems.

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If the number of equations is fewer than the number of unknowns ( under determined system ),

then the homogeneous system will have infinitely many solutions.

If the number of equations is greater or equal to the number of unknowns ( over determined

system ), then the homogeneous system will have either a unique solution or infinitely many solutions.

A homogeneous systems of linear equations is consistent. The solution set is either a

singleton or infinite.

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Every linear combination of solutions of a homogeneous system Ax=0 is again a solution of the system.

Nonhomogeneous Systems

Consider the nonhomogeneous system of m equations with n unknowns:

+ ........ + =

+ ........ + =

+ ........ + =

.............

+ ........ + =

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Every system of linear equations Ax = b has either no solution, exactly one solution or

infinitely many solutions.

Example 4: Consider the system:

> eq1:=x[1]+x[2]+2*x[3]=1: eq2:=2*x[1]+2*x[2]-4*x[3]=2:

> print(eq1); print(eq2);

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What is the solution set of this system?

> solveqns({eq1,eq2},{x[1],x[2],x[3]});

`Select one of the following modes`

`     1. Demonstration mode`

`     2. No-Step mode`

2;

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This system has infinitely many solutions with one free variable. The solution of the

nonhomogeneous system can be expressed as

> matrix(3,1,[x[1],x[2],x[3]]) = matrix(3,1,[1,0,0]) + a*matrix(3,1,[-1,1,0]);

where is the free variable.The solution [1,0,0] is called the particular solution to the system; while is a solution of the associated homogeneous system.

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Example 5: Consider the system:

> eq1:=x[1]+2*x[2]-5*x[3]+6*x[4]=10: print(eq1);

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What is the solution set of this system?

> solveqns(eq1,{x[1],x[2],x[3],x[4]});

`Select one of the following modes`

`     1. Demonstration mode`

`     2. No-Step mode`

2;

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This system has infinitely many solutions with three free variables . The general solution

[ ] of the nonhomogeneous system is

=

=

The solution [10,0,0,0] is the particular solution of the system; while the solutions

; ; and are solutions to the associated

homogeneous system.

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Let Ax=b be a linear system. If p is any particular solution of Ax=b and h is a solution of the

corresponding homogeneous system Ax=0 , then p+h is a solution of Ax=b . Moreover, every

solution of Ax=b has tghe form p+h, so that the general solution is x=p+h where Ah=0 .

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Subspaces

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A subset W of is closed under vector addition if for all u,v in W the sum u+v is in W .

If r v is in W for all v in W and all scalars r, then W is closed under scalar multiplication.

A non-empty subset W of that is closed under both vector addition and scalar

multiplication is a subspace of .

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Example 6:

> `S1 ={[x,y,z] | 2*x+3*y+z=10}`;

Choose two vectors in :

> v:=vector([-4,3,9]);u:=vector([-7,4,12]);

Compute their sum

> `v+u`=evalm(v+u);

Does vector v + u belong to subset ? Why?

Example 7:

> `S2 ={[x,y,z] | 2*x+3*y+z=0}`;

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Is subset closed under addition? under scalar multiplication?

Subset is closed under the two operations. Hence is a subspace of .

> with(plots):

```Warning, the name changecoords has been redefined
```

> implicitplot3d({2*x+3*y+z=0,2*x+3*y+z=10},x=-5..5,y=-5..5,z=0..10,axes=normal);

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We observe that the graph of passes through the origin while the graph of does not.

The zero vector belongs to the set but does not belong to the set .

Graphically, can you tell which subset does not contain the zero vector?

What is the implication of not having the zero vector in the set?

Example 8: Consider the following set

> `S1 = {[x,y] | x^2 + y^2 <= 1}`;

Note that the set {0} and the set are subspaces of .

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Let W = sp{ , . . ., } be the span of k vectors in . Then W is a subspace

of .

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Example 9: Consider the homogeneous system of linear equations

> eq1:=x1 + 3*x2 + 5*x3 = 0:
eq2:=3*x1+ 2*x2 + 5*x3 = 0:

> print(eq1); print(eq2);

The system is defined by matrix multiplication A*x = 0 where A and x are

> A:=matrix([[1,3,5],[3,2,5]]);
x:=matrix([[x1],[x2],[x3]]);

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which is equivalent to

> multiply(A,x)=0;

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Upon solving the system

> solve({eq1,eq2},{x1,x2,x3});

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we obtain

> `S ={[x1,x2,x3]|x1=(-5/7)*a,x2=(-10/7)*a,x3=a}`;

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or

> S = span(matrix(3,1,[-5/7,-10/7,1]));

or

> S = span(matrix(3,1,[-5,-10,7]));

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Is S a subspace of ?

Geometrically, S represents a line through the origin. The set S is closed under addition and

scalar multiplication. Therefore, S is a subspace. Hence, the solution set of a homogeneous

system is a subspace. This is called the nullspace of the matrix A and is denoted by N(A).

Is the solution set of a nonhomogeneous system of linear equations a subspace?

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The ROW SPACE of an nxm matrix A is the space spanned by the rows of the matrix A.

Since the span of a set of vectors is a subspace, the row space of an nxm matrix A is a

subspace of .

The COLUMN SPACE of an nxm matrix A is the space spanned by the columns of the

matrix A.

Since the span of a set of vectors is a subspace, the column space of an nxm matrix A is a

The NULL SPACE of an nxm matrix A is the solution set to Ax=0.

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Example 10: Find the row space, column space, and the null space.

> A:=matrix([[1,8,3],[6,-3,7]]);

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> AUG := matrix([[1,8,3,0],[6,-3,7,0]]);

> solve({x1+8*x2+3*x3=0,6*x1-3*x2+7*x3=0},{x1,x2,x3});

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A linear system Ax=b has a solution if and only if b is in the column space of A.

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Bases

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Let W be a subpsace of . A subset { ,. . ., } of W is a basis for W if

every vector in W can be expressed uniquely as a linear combination of ,. . ., .

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Example 11: Consider the two independent vectors

> e1:=vector([1,0]); e2:=vector([0,1]);

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Is every vector in a linear combination of the vectors and ?

Let w be any vector in

> w:= vector([x,y]);

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If w = [x,y] is a linear combination of and

> evalm(w)=c1*evalm(e1)+c2*evalm(e2);

Then = x and = y. Hence w is

> evalm(w)=x*evalm(e1)+y*evalm(e2);

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Therefore every vector in can be written uniquely as a linear combination of the two

vectors and . Thus the set { , } is spans and is a basis for .

Example 12: Now consider the vectors

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> e1:=vector([1,0]);e2:=vector([0,1]);
u:=vector([2,5]);

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Is the set { , , u} a basis for ? Let w be any vector in

> w:=vector([x,y]);

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If w = [x , y] is a linear combination of , and u

> evalm(w)=c1*evalm(e1)+c2*evalm(e2)+c3*evalm(u);

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This yields the system of equations

> eq1:=c1+2*c3=x; eq2:=c2+5*c3=y;

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On solving the system we get

> solve({eq1,eq2},{c1,c2,c3});

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Since the system is consistent, the vectors , , and u do span . However,

the linear combination is not unique.

> evalm(w)=(x-2)*evalm(e1)+(y-5)*evalm(e2)+evalm(u);

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The set { ,. . . , } is a basis for W = sp{ ,. . . , } in

if and only if the zero vector is a unique linear combination of the that is

if and only if + . . . + = 0 implies that

= = . . . = = 0.

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We call the set { , . . . , } the standard basis for .

Example 13: Consider the vectors

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> v1:=vector([0,1,1]); v2:=vector([1,1,2]);v3:=vector([1,0,2]);

> eq1:=c2 + c3=0: eq2:=c1+ c2=0: eq3:= c1+2*c2+2*c3=0:

> print(eq1); print(eq2); print(eq3);

> solve({eq1,eq2,eq3},{c1,c2,c3});

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Change vector v1

> v1 := vector([3,-2,6]);

> eq1:= 3*c1 + c2 + c3=0: eq2:=-2*c1+ c2=0: eq3:= 6*c1+2*c2+2*c3=0:

> print(eq1); print(eq2); print(eq3);

> solve({eq1,eq2,eq3},{c1,c2,c3});

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Unique Solutions for Ax=b

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Let A be a square matrix, i.e. A is a n x n matrix. The following are equivalent:

1.) The linear system Ax=b has a unique solution

2.) The matrix A is row equivalent to the identity matrix I.

3.) The matrix A is invertible.

4.) The column vectors of A form a basis for

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Let A be an m x n matrix. The following are equivalent.

1.) Each consistent system Ax=b has a unique solution

2.) The reduced row-echelon form of A consists of the n x n identity matrix

followed by m-n rows of zeros

3.) The columns of A form a basis for the column space A .

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Exercises

1, 3, 9, 12, 14, 17, 19, 23, 25, 29, 30,31, 38.